let N be a positive interger less than 10^2002. when the digit 1 is
placed after the last digit of N, the number formed is three times the
number formed when the digit 1 is placed infront of N. How many
different values of N are there???
but it's even harder without a calculator...i had to do this for a maths
challenge in school, witho...
I think the answer is 333, and I did not use a calculator, just pen and paper. Here's my
3(10^x + N) = 10N + 1
=> 7N = 3*10^x - 1
So 3*10^x -1 must be divisible by 7.
As it turns out, 3*10^x - 1 is divisible by 7 if and only if x is of the form 6y - 1. I worked
this out by calculating 3*10^n modulo 7 for n = 1 to 5; 10^6 = 1 (mod 7) (by Fermat's Little
) so x can be considered modulo 6. Hence the result, since for n=1 to 5, only
3*10^5 = 1 (mod 7).
Anyway, there are 333 such numbers <= 2002 (3*10^2003 -1 > 7*10^2002, so that's not
If you've haven't done modular arithmetic, what it means for two numbers to be congruent
(mod n) is that they differ only by a multiple of n. This is pretty nifty, as it means you can
throw away multiples of n all over the place.